PAS MATEMATIKA BLOG

Operasi Polinom

$\color{blue}\textrm{C. Operasi Pada Polinom}$

$\textbf{1. Kesamaan dua buah polinom}$

Dua buah polinom dikatakan sama jika keduanya memiliki pangkat/derajat sama dan koefisien-koefisien suku yang sejenis juga sama.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{aligned}1.\quad \textrm{Misalkan diketahui}\: \quad&\\\color{red}x^{4}+Ax^{3}-4x^{2}-10x+3&=\color{red}(x^{2}+2x+3)(x^{2}+Bx+1)\\ x^{4}+Ax^{3}-4x^{2}-10x+3&=x^{4}+(B+2)x^{3}+(2B+4)x^{2}\\ &+(3B+2)x+3\\ \textrm{Elemen yang bersesuaian}&\\ \textrm{untuk}\: \: x^{1}\: :\: \color{red}-10&=\color{red}3B+2\\ \textrm{maka}\: \: \: B& =4\\ \textrm{untuk}\: \: x^{3}\: :\: \color{red}A&=\color{red}B+2\\ A&=-2 \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: m\: \: \textrm{dan}\: \: n,\: \textrm{jika diketahui}\\ &\displaystyle \color{red}\frac{m}{x+1}+\frac{n}{x-2}=\frac{3x+4}{x^{2}-x-2}\\\\ &\textbf{Jawab}:\\ &\textrm{Kalikan kedua ruas dengan}\\ &\color{red}x^{2}-x-2\: \: \color{black}\textrm{atau}\: \: \color{red}(x+1)(x-2)\\ &\textrm{maka}\\ &\color{red}3x+4=m(x-2)+n(x+1)\\ &\Leftrightarrow 3x+4=(m+n)x+(-2m+n)\\ &\textrm{Dari bentuk kesamaan di atas didapatkan}\\ &\color{red}m+n=3\\ &\color{red}-2m+n=4\\ &\textrm{Dengan eliminasi substitusi akan}\\ &\textrm{didapatkan nilai}\: \: m=-\displaystyle \frac{1}{3}\: \: \textrm{dan}\: \: n=\frac{10}{3} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui kesamaan dua polinom}\\ &\displaystyle \color{red}5x^{2}-2x+3=ax^{2}+(b+c)x+7(b-c)\\ &\textrm{Tentukan nilai}\: \: a+8b-6c\\\\ &\textbf{Jawab}:\\ &\textrm{Dari soal diketahui bahwa}\\ &\begin{cases} a &=5 \\ b+c &=-2 \\ 7(b-c) &=3 \end{cases}\\ &\textrm{maka}\\ &7b+7c=-14\\ &7b-7c=3\qquad +\\\hline &14b=-11\Rightarrow b=-11/14\\ &\textrm{dan}\\ &7b+7c=-14\\ &7b-7c=3\qquad -\\\hline &14c=-17\Rightarrow c=-17/14\\ &\textrm{maka nilai}\: \: a+8b-6c\\ &=\color{red}5+8\left ( \displaystyle \frac{-11}{14} \right )-6\left (-\displaystyle \frac{17}{14} \right )\\ &=\color{red}5+\displaystyle \frac{14}{14}\color{black}=\color{red}5+1\color{black}=\color{red}6 \end{array}$.

$\textbf{2. Penjumlahan}$

Dua polinom dapat dijumlahkan jika hanya jika suku-sukunya sejenis, jika tidak maka tidak bisa

$\textbf{3. Pengurangan}$

Pada operasi pengurangan juga juga berlaku seperti pada operasi penjumlahan, yaitu pengurangan hanya bisa terjadi pada suku-suku yang sejenis saja yang lainnya tidak dapat dilakukan.

$\textbf{4. Perkalian}$

Pada jenis operasi ini dilakukan seperti mengalikan biasa yaitu mengalikan semua suku-suku secara distribusi dari kedua polinom tersebut.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui 2 suku banyak berikut}\\ &\begin{cases} p(x) &=x^{3}+2x^{2}+x-1 \\ q(x) &=x^{4}+5x+2 \end{cases}\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{penjumlahan keduanya}\\ &\textrm{b}.\quad \textrm{pengurangan}\: \: p(x)\: \: \textrm{oleh}\: \: q(x)\\\\ &\textrm{Jawab}:\\ &\begin{array}{lllllllllll}\\ p(x)=&&x^{3}&+&2x^{2}&+&x&-&1&\\ q(x)=&x^{4}&&&&+&5x&+&2&(+)\\\hline &\color{red}x^{4}\: +&\color{red}x^{3}&+&\color{red}2x^{2}&+&\color{red}6x&+&\color{red}1& \end{array}\\ &\textrm{poin b Silahkan dicoba sebagai latihan} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah hasil kali perkalian}\\ &\textrm{dari dua polinom berikut}\\ &\textrm{a}.\quad 3x(-5x^{2})\\ &\textrm{b}.\quad 2a(7a-3)\\ &\textrm{c}.\quad (x+2)(x-5)\\ &\textrm{d}.\quad (3t-2)(2t^{2}-5t+3)\\ &\textrm{e}.\quad (5a^{2}+2)(5a^{2}-2)\\ &\textrm{f}.\quad (x^{3}-2x)(x^{2}+3x-4)\\ &\textrm{g}.\quad (2a^{3}+1)(-a-3)^{2}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&3x(-5x^{2})=-3.5x^{1+2}=\color{red}-15x^{3}\\ \textrm{b}.\quad&2a(7a-3)=2.7a^{1+1}-2.3a=\color{red}14a^{2}-6a\\ &\color{blue}\textrm{Selanjutnya kita langsungkan saja}\\ \textrm{c}.\quad&(x+2)(x-5)=x^{2}+(2-5)x-2.5\\ &\qquad\qquad \qquad\: =\color{red}x^{2}-3x-10\\ \textrm{d}.\quad&(3t-2)(2t^{2}-5t+3)\\ &\qquad = 6t^{3}-15t^{2}+9t-4t^{2}+10t-6\\ &\qquad = \color{red}6t^{3}-19t^{2}+19t-6\\ \textrm{e}.\quad&(5a^{2}+2)(5a^{2}-2)\\ &\qquad = 25a^{4}-10x^{2}+10a^{2}-4\\ &\qquad =\color{red}25a^{4}-4\\ \textrm{f}.\quad&(x^{3}-2x)(x^{2}+3x-4)\\ &x^{5}+3x^{4}-4x^{3}-2x^{3}-6x^{2}+8x\\ &\qquad =\color{red}x^{5}+3x^{4}-6x^{3}-6x^{2}+8x\\ \textrm{g}.\quad&(2a^{2}+1)(-a-3)^{2}\\ &\qquad =(2a^{2}+1)(a^{2}+6a+9)\\ &\qquad =2a^{4}+12a^{3}+18a^{2}+a^{2}+6a+9\\ &\qquad =\color{red}2a^{4}+12a^{3}+19a^{2}+6a+9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah hasil dari perkalian}\\ &\textrm{dua polinom berikut}\\ &\textrm{a}.\quad \begin{cases} p(x) &=x^{2}-x-1 \\ q(x) &=x^{2}+x+1 \end{cases}\\\\ &\textrm{b}.\quad \begin{cases} p(x) &=x^{5}+3x^{3}-x-1 \\ q(x) &=x^{4}+2x+1 \end{cases}\\\\ &\textrm{c}.\quad \begin{cases} p(x) &=x^{6}+3x-6 \\ q(x) &=x^{3}-6x+3 \end{cases}\\\\ &\textrm{d}.\quad \begin{cases} p(x) &=x^{2020}-x \\ q(x) &=x^{2}+x-1 \end{cases}\\\\ &\textrm{e}.\quad \begin{cases} p(x) &=x^{2021}-1 \\ q(x) &=x^{2019}+1 \end{cases}\\\\ &\textrm{Jawab}:\\ &\textrm{Poin a sampai d silahkan dicoba}\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Diketahui bahwa}\\ &\begin{cases} p(x) &=x^{2021}-1 \\ q(x) &=x^{2019}+1 \end{cases}\\ &\textrm{maka}\: \: p(x)\times q(x)\\ &=\left ( x^{2021}-1 \right )\times \left ( x^{2019}+1 \right )\\ &=x^{2021+2019}+1\times x^{2021}-1\times x2019-1\times 1\\ &=\color{red}x^{4040}+x^{2021}-x^{2019}-1 \end{aligned} \end{array}$

$\textbf{5. Pembagian}$

Perhatikanlah ilustrasi pembagian bersusun panjang berikut

Misalkan untuk pembagian $x^{3}+4x^{2}-2x+4$ oleh $x-1$ adalah sebagai berikut:

Selanjutnya dari caontoh di atas kita mendapatkan,

$\begin{aligned}x^{3}&+4x^{2}-2x+4\\ &=(x-1)(x^{2}+5x+3)+7 \end{aligned}$

Sehingga dari uraian di atas secara umum pembagian polinom dapat dinyatakan bahwa:

$\textrm{Polinomial}=\textrm{Pembagi}\times \textrm{Hasil bagi}+\textrm{Sisa}$

$\textbf{a. Pembagian bentuk}\: (x-h)$

$\textbf{b. Pembagian bentuk}\: (ax+b)$

$\textbf{c. Pembagian bentuk}\: (ax^{2}+bx+c)$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah hasil bagi dan sisa pembagian}\\ &\textrm{jika}\: \: x^{3}+4x^{2}-2x+4\: \: \textrm{oleh}\: \: x-1\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah hasil bagi dan sisa pembagian}\\ &\textrm{jika}\: \: 4x^{3}-8x^{2}-x+5\: \: \textrm{oleh}\: \: 2x-1\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah hasil bagi dan sisa pembagian}\\ &\textrm{jika}\: \: x^{4}-2x^{2}-13x-19\: \: \textrm{oleh}\: \: x^{2}-2x-3\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$

Catatan hasil bagi adalah pada contoh no.1 s.d 3 adalah pada tiap pembahasan di tiap nomornya adalah terletak di bagian atas (berwarna biru) dan sisa pembagiannya adalah yang terletak di bagian paling bawah (berwarna merah).

DAFTAR PUSTAKA

Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Sukino. 2017. Matematika Jilid 2 untuk SMA/MA Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Polinom (Suku Banyak)

$\color{blue}\textrm{A. Pendahuluan}$

Polinom disebut juga suku banyak. Polinom atau suku banyak adalah suatu bentuk variabel yang berpangkat/berderajat.

Secara definisi suku banyak (polinomial) dalam $x$ berderajat $n$ adalah:

Suatu bentuk

$\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}$

dengan $n$ bilangan cacah serta $a_{0},\: a_{1},\: a_{2},\: ...,\: a_{n}$ koefisien dari suku $x$ dan $a_{n}\neq 0$ dengan $a_{0}$ sebagai suku tetap (konstan)nya.

Selanjutnya perhatikanlah tabel berikut!

$\color{purple}\begin{array}{|l|l|}\hline \begin{aligned}a_{n}&\: \: \textrm{adalah koefisien dari} \: \: x^{n}\\ a_{n-1}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-1}\\ a_{n-2}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-2}\\ \vdots &\\ a_{2}&\: \: \textrm{adalah koefisien dari} \: \: x^{2}\\ a_{1}&\: \: \textrm{adalah koefisien dari} \: \: x^{1}\\ a_{0}&\: \: \textrm{adalah konstanta} \\ &(\textrm{suku tetap}) \end{aligned}&\begin{aligned}a_{n}\: &\: \neq 0\\ n:&\: \: \textrm{bilangan cacah},\\ :&\: \: \textrm{adalah derajat (pangkat)} \\ &\: \: \textrm{tertinggi dalam suku} \\ &\: \: \textrm{banyak tersebut}&\\ &\\ &\\ &\end{aligned}\\\hline \end{array}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 1}$

$\begin{aligned}1.\quad&\textrm{Polinom}\: \: \color{red}2x^{3}-6x^{2}+2020\: \: \color{black}\textrm{dapat dinyatakan}\\ &\textrm{dengan}\: \: \: \color{blue}2x^{3}-6x^{2}+0x^{1}+2020x^{0}\\ &\textrm{Polinom tersebut memiliki suku tetap}\: \: 2020\\ 2.\quad&\textrm{Polinom}\: \: \color{red}5x^{4}-8x^{3}+6x-2021 \: \: \color{black}\textrm{dapat dinyatakan}\\ &\textrm{dengan}\: \: \: \color{blue}5x^{4}-8x^{3}+0x^{2}+6x^{1}-2021x^{0}\\ &\textrm{Polinom tersebut memiliki suku tetap}\: \: -2021\\ 3.\quad&\textrm{Polinom}\: \: \color{red}x^{4}-2x^{3}+3x^{2}-2\sqrt{x}+1 \: \: \color{black}\textrm{tidak dapat}\\ &\textrm{dinamakan polinom, sebab ada variabel dari}\: \: \: \color{blue}x\\ &\textrm{yang berderajat bukan bilangan cacah}\\ 4.\quad&\textrm{Sedangkan polinom}\: \: \color{red}5-x+(2-x)(1+x+x^{2})\\ &\textrm{adalah bentuk polinom, karena dapat dinayatakan}\\ &\textrm{dengan}\: \: \: \color{blue}-x^{3}+x^{2}+7 \end{aligned}$

$\color{blue}\textrm{B. Nilai Polinom}$

Polinom atau suku banyak yang berderajat $\color{red}n$ yang selanjutnya dinyatakan dengan

$f(x)=\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{1}x^{1}+a_{0}$

Berkaitan dengan kebutuhan penentuan nilai ini, dapat ditentukan dengan dua cara:

$\textbf{a. Substitusi}$

$\begin{aligned}&\textrm{Nilai suku banyak}\: \: \color{red}f(x)\: \: \textrm{berderajat}\\ &n\: \: \textrm{saat}\: \: \color{red}x = k\: \: \color{black}\textrm{adalah}\: \: \color{blue}f(k).\\ &\textrm{Jika}\: \: f(k)=0\: \: \textrm{maka}\: \: x = k\: \: \textrm{akar dari}\: \: f(x),\\ &\textrm{dan}\: \: (x-k)\: \: \textrm{faktor dari}\: \: f(x)\\ &\end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 2}$

Jika suatu polinom dinyatakan dengan $f(x)$, maka nilai polinom itu untuk $x=3$ adalah $f(3)$.

Misalkan diketahui

$\begin{aligned}1.\quad f(x)&=x^{3}-1\\ \textrm{mak}&\textrm{a}\\ f(1)&=1^{3}-1=1-1=0\\ f(3)&=3^{3}-1=27-1=26\\ f(-4)&=(-4)^{2}-1=-64-1=-65 \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: h(x)=2x^{3}+5x^{2}-12x-6\\ &\textrm{Tentukanlah nilai untuk}\: \: h(-2),\: h(-1),\\ &h(0),\: h(1),\: \: \textrm{dan}\: \: h(2)\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{|c|c|l|}\hline \color{red}x=k&\color{red}h(k)&\qquad\qquad\qquad\qquad\color{red}\textrm{Nilai}\\\hline x=-2&h(-2)&\begin{aligned}h(-2)&=2(-2)^{3}+5(-2)^{2}-12(-2)-6\\ &=-16+20+24-6\\ &=22 \end{aligned}\\\hline x=-1&h(-1)&\begin{aligned}h(-1)&=2(-1)^{3}+5(-1)^{2}-12(-1)-6\\ &=-2+5+12-6\\ &=9 \end{aligned}\\\hline x=0&h(0)&\begin{aligned}h(0)&=2(0)^{3}+5(0)^{2}-12(0)-6\\ &=-6 \end{aligned}\\\hline x=1&h(1)&\begin{aligned}h(1)&=2(1)^{3}+5(1)^{2}-12(1)-6\\ &=2+5-12-6\\ &=-11 \end{aligned}\\\hline x=2&h(2)&\begin{aligned}h(2)&=2(2)^{3}+5(2)^{2}-12(2)-6\\ &=16+20-24-6\\ &=6 \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: p(x)=x-2019\\ &\textrm{dan}\: \: q(x)=x^{2019}+1.\: \textrm{Tentukanlah}\\ &\textrm{nilai untuk}\: \: p\left ( q(2) \right )\: \: \textrm{dan}\: \: q\left ( p(2) \right )\\\\ &\color{blue}\textrm{Jawab}:\\ &\textrm{Yang dibahas yang bagian}\: \: p\left ( q(2) \right )\\ &q(2)=2^{2019}+1,\: \textrm{maka nilai}\\ &\begin{aligned}p\left ( q(2) \right )&=\left ( 2^{2019}+1 \right )-2019\\ &=2^{2019}-2018 \end{aligned}\\\\ &\textrm{Untuk yang}\: \: q\left ( p(2) \right )\: \: \textrm{adalah}\\ &p(2)=\cdots , \: \textrm{maka nilai}\\ &\begin{aligned}q\left ( p(2) \right )&=\because \cdots ^{2019}+1\\ &=\cdots \end{aligned} \end{array}$

$\textbf{b. Horner/Sintetik}$

Nilai suatu polinom dapat ditentukan dengan pembagian sintesis Horner

Misalkan:

$\begin{aligned}f(x)&=\color{blue}ax^{3}+bx^{2}+cx+d\: \: \color{black}\textrm{saat akan dibagi}\\ &\color{red}x=h,\: \: \color{black}\textrm{maka pembagian Horner itu}:\\ & \end{aligned}$

Perhatikan bahwa proses ke bawah adalah berup proses penjumlahan.

Proses di atas akan sama saat kita mensubstitusikan $\color{red}x=h$ ke dalam $\color{red}f(x)$, yaitu:

$\begin{aligned}f(x)&=\color{blue}ax^{3}+bx^{2}+cx+d\: \: \textrm{saat}\\ &\color{red}x=h,\: \: \color{black}\textrm{maka}\\ f(\color{red}h\color{black})&=a\color{red}h^{3}\color{black}+b\color{red}h^{2}\color{black}+c\color{red}h\color{black}+d\\ &\\ &\textbf{Cukup JELAS bukan}? \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 3}$

$\begin{array}{l}\\ \textrm{Tentukanlah nilai dari}\: \: f(4)\: \: \textrm{jika}\\ \textrm{diketahui}\: \: f(x)=x^{3}-x-5\\ \textrm{Jawab}:\\ \begin{aligned}(1).\quad&\textrm{Cara substitusi langsung}\\ &f(x)=x^{3}-x-5\\ &f(4)=\color{red}4^{3}-4-5\\ &\qquad=\color{red}64-9=\color{blue}55\\ (2).\quad&\textrm{Cara Horner}\\ &\textrm{Karena}\: \: f(x)=x^{3}-x-5\\ &\textrm{dan koefisiennya yang akan}\\ &\textrm{adalah}:\\ & a_{3}=1,\: a_{2}=0,\: a_{1}=-1,\: \&\: a_{0}=-5\\ &\textbf{maka bagan pembagian Hornernya}\\ &\begin{array}{ll|llllllllll}\\ &\color{red}x=4&1&\color{blue}0&\color{magenta}-1&-5&\\ &&&&&&\\ &&&\color{blue}4&\color{magenta}16&60&+\\\hline &&1&\color{blue}4&\color{magenta}15&55 \end{array} \end{aligned} \end{array}$

Contoh 8 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 36.&\textrm{Persamaan lingkaran yang menyinggung}\\ &\textrm{sumbu X serta melalui titik potong}\\ &\textrm{lingkaran}\: \: (x+1)^{2}+(y+2)^{2}=1\: \: \textrm{dan}\\ &x^{2}+y^{2}+3x+3y+4=0\: \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-4x+2y+4=0\\ &\textrm{b}.\quad x^{2}+y^{2}-4x+2y-4=0\\ &\textrm{c}.\quad x^{2}+y^{2}-4x-2y-4=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}+4x+2y+4=0\\ &\textrm{e}.\quad x^{2}+y^{2}+4x+2y-4=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=(x+1)^{2}+(y+2)^{2}=1\\ &\qquad \Leftrightarrow x^{2}+y^{2}+2x+4y+4=0\\ &\bullet \: L_{2}=x^{2}+y^{2}+3x+3y+4=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=-x+y=0\Leftrightarrow y=x\\ &\color{blue}\textrm{Dengan cara coba-coba, maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}+2x+4y+4+p(-x+y)=0\\ &\color{blue}\textrm{Untuk}\: \: p=1\\ &\Leftrightarrow x^{2}+y^{2}+2x+4y+4+(-x+y)=0\\ &\Leftrightarrow x^{2}+y^{2}+x+5y+4=0\\ &\color{blue}\textrm{Untuk}\: \: p=-1\\ &\Leftrightarrow x^{2}+y^{2}+2x+4y+4-(-x+y)=0\\ &\Leftrightarrow x^{2}+y^{2}+3x+3y+4=0\\ &\color{blue}\textrm{Dan untuk}\: \: p=-2\\ &\Leftrightarrow x^{2}+y^{2}+2x+4y+4-2(-x+y)=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}+4x+2y+4=0 \end{aligned} \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnyanya} \end{array}$

DAFTAR PUSTAKA

Kartini, Suprapto, Subandi, dan Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 7 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 31.&\textrm{Persamaan lingkaran yang melalui titik}\\ &(0,0)\: \: \textrm{dan titik potong kedua lingkaran}\\ &x^{2}+y^{2}-6x-8y-11=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-4x-6y-22=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-12x+10y=0\\ &\textrm{b}.\quad x^{2}+y^{2}+8x-10y=0\\ &\textrm{c}.\quad x^{2}+y^{2}-8x+12y=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}-8x-10y=0\\ &\textrm{e}.\quad x^{2}+y^{2}+12x-8y=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-6x-8y-11=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-4x-6y-22=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=-2x-2y+11=0\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (0,0), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-6x-8y-11 +p(-2x-2y+11)=0\\ &\Leftrightarrow 0^{2}+0^{2}-0-0-11+p(0+11)=0\\ &\Leftrightarrow p=\color{blue}1 \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-6x-8y-11+(-2x-2y+11)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-8x-10y=0 \end{aligned} \end{array}$.

Berikut ilustrasi gambarnya

$\begin{array}{ll}\\ 32.&\textrm{Persamaan lingkaran yang melalui titik}\\ & (8,4)\: \: \textrm{dan titik potong lingkaran}\: x^{2}+y^{2}=16\\ &\textrm{dan}\: \: x^{2}+y^{2}-4x-4y=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-8x-8y-16=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x+8y+16=0\\ &\textrm{c}.\quad \color{red}x^{2}+y^{2}-8x-8y+16=0\\ &\textrm{d}.\quad x^{2}+y^{2}+8x+8y-16=0\\ &\textrm{e}.\quad x^{2}+y^{2}+8x+8y+16=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-16=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-4x-4y=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=4x+4y-16=0\\ &\Leftrightarrow x+y=4\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (8,4), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-16+p(x+y-4)=0\\ &\Leftrightarrow 8^{2}+4^{2}-16+p(8+4-4)=0\\ &\Leftrightarrow -8p=\color{blue}64\color{black}\Leftrightarrow p=\color{blue}-8 \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-16-8(x+y-4)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-8x-8y+16=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnyanya} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Persamaan lingkaran yang melalui titik}\\ & (7,-4)\: \: \textrm{dan titik potong kedua lingkaran}\\ &x^{2}+y^{2}-6x+8y-27=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-26x+4y+121=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-36x-2y+121=0\\ &\textrm{b}.\quad x^{2}+y^{2}+24x-4y-222=0\\ &\textrm{c}.\quad 3x^{2}+3y^{2}-18x+2y-121=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}-36x+2y+195=0\\ &\textrm{e}.\quad x^{2}+y^{2}+24x+2y+195=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-6x+8y-27=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-26x+4y+121=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=20x+4y-148=0\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (7,-4), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-6x+8y-27\\ &\qquad+p(20x+4y-148)=0\\ &\Leftrightarrow 7^{2}+(-4)^{2}-42-32-27\\ &\qquad+p(140-16-148)=0\\ &\Leftrightarrow -24p=\color{blue}36\color{black}\Leftrightarrow p=\color{blue}-\displaystyle \frac{3}{2} \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-6x+8y-27\\ &\qquad-\displaystyle \frac{3}{2}(20x+4y-148)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-36x+2y+195=0 \end{aligned} \end{array}$.

Berikut ilustrasi gambarnya

Jika dimensi gambar diperkecil menjadi

$\begin{array}{ll}\\ 34.&\textrm{Persamaan lingkaran yang melalui perpotongan}\\&\textrm{lingkaran}\: \: x^{2}+y^{2}-12x+6y+20=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-16x-14y+64=0\: \: \textrm{serta pusatnya}\\ &\textrm{terletak pada garis}\: \: 8x-3y-19=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-20x-34y+108=0\\ &\textrm{b}.\quad x^{2}+y^{2}-16x+12y+96=0\\ &\textrm{c}.\quad x^{2}+y^{2}-12x+20y+88=0\\ &\textrm{d}.\quad x^{2}+y^{2}+16x-24y+108=0\\ &\textrm{e}.\quad x^{2}+y^{2}+22x-34y+96=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa persamaan lingkaran}:\\ &\bullet \: L_{1}=x^{2}+y^{2}-12x+6y+20=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-16x-14y+64=0\\ &\textrm{Persamaan tali busurnya (garis kuasa)}\\ &\textrm{adalah}:\\ &L_{1}(x,y)-L_{2}(x,y)\\ &=4x+20y-44=0\Leftrightarrow \color{blue}x=11-5y\\ &\textrm{Selanjutnya dengan substitusi }\\ &\begin{aligned}&x^{2}+y^{2}-12x+6y+20=0\\ &\Leftrightarrow (x-6)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow (\color{blue}11-5y\color{black}-6)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow (y-5y)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow 26y^2-44y+9=0 \end{aligned}\\ &\textrm{Sehingga dengan}\: \: \color{red}\textrm{memodifikasi}\\ &\begin{aligned}&26y^2-44y+9=0\\ &\Leftrightarrow 25y^2-44y+y^2+9=0\\ &\quad\textrm{arahkan ke bentuk kuadrat sempurna}\\ &\Leftrightarrow 25y^2-10y+1+y^2-34y+8=0\\ &\Leftrightarrow 25y^2-10y+1+y^2-34y+17^{2}-17^{2}+8=0\\ &\Leftrightarrow (5y-1)^{2}+(y-17)^{2}-281=0\\ &\quad \textrm{ingat bahwa ada tali busur}\: \: \color{blue}5y=11-x\\ &\Leftrightarrow (\color{blue}11-x\color{black}-1)^{2}+(y-17)^{2}-281=0\\ &\Leftrightarrow (10-x)^{2}+(y-17)^{2}-281=0\\ &\Leftrightarrow x^{2}-20x+100+y^{2}-34y+289-281=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-20x-34y+108=0 \end{aligned} \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Persamaan lingkaran dengan titik pusat}\\ &\textrm{pada garis}\: \: x+2y-3=0\: \: \textrm{dan melalui}\\ &\textrm{titik potong dua lingkaran}\\ &x^{2}+y^{2}-2x-4y+1=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-4x-2y+4=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-6x+7=0\\ &\textrm{b}.\quad x^{2}+y^{2}-3y+4=0\\ &\textrm{c}.\quad x^{2}+y^{2}-2x-2y+1=0\\ &\textrm{d}.\quad x^{2}+y^{2}-2x-4y+4=0\\ &\textrm{e}.\quad x^{2}+y^{2}-3x-2y+7=0\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Gunakan cara pembahasan sebagaimana pada}\\ &\textrm{nomor-nomor sebelumnya}\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}-2x-4y+1=0,\: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+y^{2}-4x-2y+4=0\\ &\textrm{Persamaan}\: \: \color{red}\textrm{tali busur}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}-2x-4y+1\\ &-(x^{2}+y^{2}-4x-2y+4)=0\\ &\Leftrightarrow 2x-2y-3=0\\ &\textrm{Selanjutnya perlu ditentukan juga}\\&\textrm{Persamaan}\: \: \color{red}\textrm{berkas lingkaran}\: \color{black}\textrm{melalui}\\ &\textrm{titik-titik potong kedua lingkaran}\\ &\textrm{di atas adalah}:\\ &L_{1}+\lambda L_{2}=0\\ &x^{2}+y^{2}-2x-4y+1\\ &\qquad+\lambda \left ( x^{2}+y^{2}-4x-2y+4 \right )=0\\ &\Leftrightarrow (1+\lambda )x^{2}+(1+\lambda )y^{2}-(2+4\lambda )x\\ &\qquad -(4+2\lambda )y+1+4\lambda =0\\ &\textrm{Saat}\: \: \lambda =-1,\: \textrm{maka persamaan berkas}\\ &\textrm{lingkarannya adalah}:\: 2x-2y-3=0\\ &\textrm{Hal ini hasilnya sama persis saat kita}\\ &\textrm{menentukan persamaan}\: \color{red}\textrm{tali busur}\: \color{black}\textrm{di atas}\\ &\textrm{Selanjutnya kita ambil}\\ &L_{2}-(L_{1}+\lambda L_{2})=0\\ &\Leftrightarrow x^{2}+y^{2}-4x-2y+4-(2x-2y-3)=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-6x+7=0 \end{aligned} \end{array}$.

Gambar mula-mula

Lingkaran baru yang berpusat di (3,0)

Invers Fungsi Trigonometri, Koordinat Kartesius dan Koordinat Kutub

Lanjutan Materi Fungsi Trigonometri dan Grafiknya

F. 2 Garfik Fungsi Trigonometri

F. 2. 1 Grafik Fungsi Sinus

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c| }\hline \color{magenta}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0\\\hline \color{magenta}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi & \\\hline \color{red}f(x)&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0&\\\hline \end{array}$.

F. 2. 2 Grafik Fungsi Cosinus

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \color{black}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1\\\hline \color{black}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi&\\\hline \color{red}f(x)&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1&\\\hline \end{array}$.

F. 2. 3 Grafik Fungsi Tangen

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \color{black}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&0&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0\\\hline \color{black}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi&\\\hline \color{red}f(x)&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0&\\\hline \end{array}$.

Pada fungsi Tangen demikian juga nanti Cotangennya ada beberapa nilai fungsinya yang tidak terdefinisi. Dalam fungsi Tangen fungsi, nilai fungsi yang tidak terdefini terdapat pada saat nilai $x=\displaystyle \frac{\pi }{2}=90^{\circ}$ dan $x=\displaystyle \frac{3\pi }{2}=270^{\circ}$. Sehingga pada saat posisi nilai itu, maka dibuatlah garis bantu berupa garis putus-putus pada grafik yang dan ditampakkan berupa garis vertikal yang selanjutnya garis vertikal itu disebut sebagai asimtot.

F. 2. 4 Menggambar Grafik Fungsi Trigonometri

$\begin{aligned}&\textrm{untuk bentuk}\\ &f(x)=\begin{cases} y &=a\sin bx+c \\ y &=a\cos bx+c \\ y & =a\tan bx+c \end{cases}\\ &\begin{array}{|c|l|l|}\hline 1.&a&\textrm{Amplitudo}\\\hline 2.&b&\textrm{Periode}\\\hline 3.&c&\textrm{Geseran}\\\hline \end{array} \end{aligned}$.

$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Gambarlah grafik fungsi berikut} \\ &\textrm{jika}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{a}.\quad f(x)=-2\sin x\\ &\textrm{b}.\quad f(x)=3\cos x\\ &\textrm{c}.\quad f(x)=\displaystyle \frac{1}{2}\sin x\\ &\textrm{d}.\quad f(x)=4\cos x\\ &\textrm{e}.\quad f(x)=2\tan x\\\\&\color{blue}\textrm{Jawab}:\\&\begin{aligned}& \end{aligned} \end{array}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 a}\\ &y=f(x)=-2\sin x=a\sin bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |-2 \right |=2\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 b}\\ &y=f(x)=3\cos x=a\cos bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |3 \right |=3\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 c}\\ &y=f(x)=\displaystyle \frac{1}{2}\sin x=a\sin bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle \frac{1}{2} \right |=\displaystyle \frac{1}{2}\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 d}\\ &y=f(x)=\displaystyle 4\cos x=a\cos bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 4 \right |=\displaystyle 4\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 e}\\ &y=f(x)=\displaystyle 2\tan x=a\tan bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 2 \right |=\displaystyle 2\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Gambarlah grafik fungsi berikut} \\ &\textrm{jika}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{a}.\quad f(x)=\left |-2\sin x \right |\\ &\textrm{b}.\quad f(x)=\left |3\cos x \right |\\ &\textrm{c}.\quad f(x)=\left |\displaystyle \frac{1}{2}\sin x \right |\\ &\textrm{d}.\quad f(x)=\left |4\cos x \right |\\ &\textrm{e}.\quad f(x)=\left |2\tan x \right |\\\\&\color{blue}\textrm{Jawab}:\\&\begin{aligned}& \end{aligned} \end{array}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.2 a}\\ &y=f(x)=\left |-\displaystyle 2\sin x \right |=\left |a\sin bx+c \right |\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 2 \right |=\displaystyle 2\\\hline 2.&b&\textrm{Periode}&\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.2 b}\\ &y=f(x)=\left |\displaystyle 3\cos x \right |=\left |a\cos bx+c \right |\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 3 \right |=\displaystyle 3\\\hline 2.&b&\textrm{Periode}&\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array} \end{aligned}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

Silahkan selesaikan soal yg belum dibahas

DAFTAR PUSTAKA

Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Fungsi Trigonometri dan Grafiknya

F. Fungsi Trigonometri dan Grafiknya

F. 1 Fungsi Trigonometri

Perhatikan ilustrasi berikut ini

Dengan

Adapun gambar dari fungsi atau pemetaan trigonometrinya dari setiap sudut $\alpha $ ke salah satu nilai dari $\sin \alpha$ , $\cos \alpha$, maupun $\tan \alpha$ dalam wilayah bilangan real adalah sebagaimana ilustrasi berikut:

$\begin{aligned}&\textrm{Misalkan}\: \: A\: \: \textrm{dan}\: \: B\: \: \textrm{dua himpunan}\\ &\textrm{Suatu relasi}\: \: F\subseteq A\times B\: \: \textrm{disebut fungsi jika}\\ &\textrm{setiap}\: a\in A,\: \textrm{maka hanya ada tepat satu}\: \: b\in B\\ &\textrm{dengan}\: \: (a,b)\in F.\\ &\textrm{Fungsi}\: \: F\: \: \textrm{disebut dengan fungsi dari}\: \: A\: \: \textrm{ke}\: \: B\\ &\textrm{Selanjutnya}\: \: A\: \: \textrm{dinamakan}\: \: \textbf{Domain}\: \: \textrm{atau}\\ &\textrm{daerah asal atau juga daerah definisi fungsi}\\ &\textrm{dan}\: \: B\: \: \textrm{disebut}\: \: \textbf{Kodomain}\\ &\textrm{Himpunan}\: \: \left \{ b\in B|(a,b)\in F \right \}\: \textrm{selanjutnya disebut}\\ &\textrm{sebagai}\: \: \textbf{nilai fungsi}\\ &\textrm{Jika}\: \: (a,b)\in F,\: \: \textrm{maka dapat tuliskan dengan}\\ &b=F(a),\: \: \textrm{yaitu nilai fungsi}\: \: F\: \: \textrm{di titik}\: \: a\\\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\: \: \quad\color{red}\textrm{Gambar}&\color{red}\textrm{Fungsi}\: \left ( f:\mathbb{R}\Rightarrow \mathbb{R} \right )\\\hline 1&\textrm{Fungsi Sinus}&\begin{aligned}&f:\alpha \Rightarrow \sin \alpha \end{aligned}\\\hline 2&\textrm{Fungsi Cosinus}&\begin{aligned}&f:\alpha \Rightarrow \cos \alpha \end{aligned}\\\hline 3&\textrm{Fungsi Tangen}&\begin{aligned}&f:\alpha \Rightarrow \tan \alpha \end{aligned}\\\hline \end{array} \end{aligned}$.

$\begin{aligned}&\textrm{Jangan lupa, sebagai pengingat kita untuk}\\ &\textrm{nilai sudut istimewanya adalah sebagai berikut}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \alpha &0^{\circ}&30&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textrm{TD}&0\\\hline \end{array} \end{aligned}$.

$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: f(x)=\sin 2x,\: \: \textrm{tentukan nilai}\\ &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi \right )\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&f(60^{\circ})=\sin 2\left ( 60^{\circ} \right )=\sin 120^{\circ}\\ &\: \: \: \qquad =\sin \left ( 180^{\circ}-60^{\circ} \right )=\sin 60^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad&f\left ( \displaystyle \frac{1}{3}\pi \right )=\sin 2\left ( \displaystyle \frac{1}{3}\pi \right )=\sin \left ( \displaystyle \frac{2}{3}\pi \right )\\ &\: \: \qquad\quad =\sin \left ( \displaystyle \frac{2}{3}(180^{\circ}) \right )=\sin 120^{\circ}=\displaystyle \frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\: \: f(x)=\sin x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{2}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{4}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&f(x)=\displaystyle \frac{1}{2}=\sin x\Rightarrow x=30^{\circ}\\ \textrm{b}.\quad&f(x)=\displaystyle \frac{1}{4}=\sin x\Rightarrow x=\color{red}\sin ^{-1}\left ( \displaystyle \frac{1}{4} \right )\\ &\textrm{hal ini dikarenakan}\: \: \displaystyle \frac{1}{4}\: \: \textrm{bukanlah}\\ &\textrm{nilai dari salah satu sudut istimewa}\\ &\textrm{untuk fungsi}\: \: \textbf{sinus} \end{aligned} \end{array}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: f(x)=\cos 2x,\: \: \textrm{tentukan nilai}\\ &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi \right )\end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\: \: f(x)=\tan x,\: \: \textrm{tentukan nilai}\\ &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi \right )\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika diketahui}\: \: f(x)=\cos x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{2}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{4} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika diketahui}\: \: f(x)=\tan x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{3}\sqrt{3}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{6}\sqrt{3} \end{array}$

DAFTAR PUSTAKA

Budhi, W.S. 2014. Bupena Matematika SMA/MA Kelas X Kelompok Wajib. Jakarta: ERLANGGA.
Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Lanjutan Identitas Trigonometri

$\begin{aligned}&\textrm{E. 3 Menentukan Nilai Perbandingan Trigonometri}\\ &\quad\textrm{pada Segitiga Siku-Siku} \end{aligned}$.

$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \tan \theta =\displaystyle \frac{a}{x} \\ &\textrm{Tentukanlah nilai}\: \: \displaystyle \frac{x}{\sqrt{a^{2}+x^{2}}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\textrm{Perhatikanlah gambar segitiga AOX berikut} \end{array}$

$.\: \qquad\begin{aligned}&\textrm{Dengan rumus Pythagoras dapatr ditentukan}\\ &\textrm{panjang ruas}\: \: \textrm{AX, yaitu}:\\ &AO^{2}+OX^{2} =AX^{2}\\ &\textrm{atau}\\ &AX^{2}=AO^{2}+OX^{2} \\ &AX=\sqrt{AO^{2}+OX^{2}}\\ &\qquad =\sqrt{x^{2}+a^{2}},\\ &\textrm{maka}\\ &\bullet \quad \sin \theta =\displaystyle \frac{a}{\sqrt{x^{2}+a^{2}}}\\ &\bullet \quad \cos \theta =\displaystyle \frac{x}{\sqrt{x^{2}+a^{2}}} \\ &\textrm{Jadi, nilai}\: \: \displaystyle \frac{x}{\sqrt{x^{2}+a^{2}}}=\color{red}\cos \theta \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: \sin \beta +\cos \beta =\displaystyle \frac{6}{5},\: \textrm{tentukanlah}\\ &\textrm{a}.\quad \sin \beta \cos \beta \\ &\textrm{b}.\quad \sin ^{3}\beta +\cos ^{3}\beta \\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\sin \beta +\cos \beta=\displaystyle \frac{6}{5}\\ &\color{red}\textrm{saat masing-masing ruas dikuadratkan,}\\ &\textrm{maka}\\ &\left (\sin \beta +\cos \beta \right )^{2}=\left (\displaystyle \frac{6}{5} \right )^{2}\\ &\sin ^{2}\beta +2\sin \beta \cos \beta +\cos ^{2}\beta =\displaystyle \frac{36}{25}\\ &\sin ^{2}\beta +\cos ^{2}\beta +2\sin \beta \cos \beta=\displaystyle \frac{36}{25}\\ &1+2\sin \beta \cos \beta=\displaystyle \frac{36}{25}\\ &2\sin \beta \cos \beta=\displaystyle \frac{36}{25}-1\\ &2\sin \beta \cos \beta=\displaystyle \frac{36-25}{25}=\frac{11}{25}\\ &\sin \beta \cos \beta=\color{blue}\displaystyle \frac{11}{50} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\sin ^{3}\beta +\cos ^{3}\beta \\ &=\left ( \sin \beta +\cos \beta \right )\left ( \sin ^{2}\beta +\cos ^{2}\beta -\sin \beta \cos \beta \right )\\ &=\left ( \sin \beta +\cos \beta \right )\left ( 1 -\sin \beta \cos \beta \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( 1-\displaystyle \frac{11}{50} \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( \displaystyle \frac{50-11}{50} \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( \displaystyle \frac{39}{50} \right )\\ &=\displaystyle \color{blue} \frac{3\times 39}{5\times 25}=\frac{117}{125} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \tan \alpha =\displaystyle \frac{1}{\sqrt{7}},\: \textrm{tentukanlah}\\ &\left ( \displaystyle \frac{\csc ^{2}\alpha -\sec ^{2}\alpha }{\csc ^{2}\alpha +\sec ^{2}\alpha} \right ) \\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \tan \alpha &=\displaystyle \frac{1}{\sqrt{7}},\: \: \color{red}\textrm{dan ingat juga bahwa}\\ \sec ^{2}\alpha &=\tan ^{2}\alpha +1=\left ( \displaystyle \frac{1}{\sqrt{7}} \right )^{2}+1=\frac{1}{7}+1=\frac{8}{7}\\ \color{red}\textrm{Demik}&\color{red}\textrm{ian juga},\: \color{black}\cot \alpha =\displaystyle \frac{1}{\tan \alpha } =\displaystyle \frac{1}{\left ( \frac{1}{\sqrt{7}} \right )}=\sqrt{7},\\ \textrm{maka},&\: \: \csc ^{2}\alpha =\cot ^{2}\alpha +1=\left ( \sqrt{7} \right )^{2}+1=7+1=8\\ \textrm{Selanj}&\textrm{utnya}\\ &\left ( \displaystyle \frac{\csc ^{2}\alpha -\sec ^{2}\alpha }{\csc ^{2}\alpha +\sec ^{2}\alpha} \right )=\left ( \displaystyle \frac{8-\displaystyle \frac{8}{7}}{8+\displaystyle \frac{8}{7}} \right )\\ &=\displaystyle \frac{\displaystyle \frac{56-8}{7}}{\displaystyle \frac{56+8}{7}} \\ &=\displaystyle \frac{48}{64}\\ &=\color{blue}\displaystyle \frac{3}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \beta\: \: \textrm{sudut lancip dan}\: \: \cos \beta =\displaystyle \frac{3}{5},\\ &\textrm{tentukan nilai dari}\: \: \displaystyle \frac{\sin \beta \tan \beta -1}{2\tan ^{2}\beta }\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\\ \cos \beta &=\displaystyle \frac{3}{5}\Rightarrow \sin ^{2}\beta +\cos ^{2}\beta =1\\ \sin ^{2}\beta &+\cos ^{2}\beta =1\\ \sin \beta &=\sqrt{1-\cos ^{2}\beta}=\sqrt{1-\left ( \displaystyle \frac{3}{5} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{9}{25}}=\sqrt{\displaystyle \frac{16}{25}}=\displaystyle \frac{4}{5}\\ \textrm{Sehingga}\: &\tan \beta =\displaystyle \frac{\sin \beta }{\cos \beta }=\frac{\displaystyle \frac{4}{5}}{\displaystyle \frac{3}{5}}=\frac{4}{3}\\ &\color{red}\displaystyle \frac{\sin \beta \tan \beta -1}{2\tan ^{2}\beta }\color{black}=\displaystyle \frac{\displaystyle \frac{4}{5}\times \frac{4}{3}-1}{2\left ( \displaystyle \frac{4}{3} \right )^{2}}\\ &\: \, \quad\quad\quad\quad\quad\quad =\displaystyle \frac{\displaystyle \frac{16}{15}-1}{\displaystyle \frac{32}{9}}=\displaystyle \frac{\displaystyle \frac{1}{15}}{\displaystyle \frac{32}{9}}=\displaystyle \frac{9}{32\times 15}\\ &\: \, \quad\quad\quad\quad\quad\quad =\displaystyle \frac{3}{32\times 5}\\ &\: \, \quad\quad\quad\quad\quad\quad =\color{blue}\displaystyle \frac{3}{160} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Identitas Trigonometri

$\Large\textrm{E Identitas Trigonometri}$.

E. 1 Nilai Trigonometri Sudut

$\textrm{a. Perbandingan Trigonometri dalam Segitiga Siku-Siku}$.

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{matrix} \bullet \quad \sin 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \cos 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \tan 45^{\circ}=1 \qquad\qquad\: \: \end{matrix}$.

$\begin{matrix} \bullet \quad \csc 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \sec 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \cot 45^{\circ}=1 \: \: \, \end{matrix}$.

Berikut ilustrasi segitiga dengan sudut istimewa yang lain yaitu $30^{\circ}$ dan $60^{\circ}$.

$\begin{array}{|c|c|}\hline \begin{matrix} \bullet \quad \color{purple}\sin 30^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \cos 30^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3} \: \: \,\\ \bullet \quad \color{blue}\sin 60^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \cos 60^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \sqrt{3}\\ \end{matrix} &\begin{matrix} \bullet \quad \csc 30^{\circ}=\displaystyle 2\\ \bullet \quad \sec 30^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \cot 30^{\circ}=\displaystyle \sqrt{3} \: \: \,\\ \bullet \quad \color{red}\csc 60^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \sec 60^{\circ}=\displaystyle 2\\ \bullet \quad \color{purple}\cot 30^{\circ}=\displaystyle \frac{1}{3}\sqrt{3}\\ \end{matrix} \\\hline \end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{aligned}\textrm{a}.\quad&\color{purple}\sin \alpha =\displaystyle \frac{y}{r}\\ \textrm{b}.\quad&\cos \alpha =\displaystyle \frac{x}{r}\\ \textrm{c}.\quad&\color{blue}\tan \alpha =\displaystyle \frac{y}{x}\\ \textrm{d}.\quad&\csc \alpha =\displaystyle \frac{r}{y}\\ \textrm{e}.\quad&\sec \alpha =\displaystyle \frac{r}{x}\\ \textrm{f}.\quad&\color{red}\cot \alpha =\displaystyle \frac{x}{y}\\ \end{aligned}$.

E. 2 Identitas Trigonometri Dasar

$\textrm{a. Dalil Pythagoras Segitiga Siku-Siku}$.

$\begin{array}{c|c}\\ \begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\color{red}\textrm{atau}\\ &c=\sqrt{a^{2}+b^{2}} \end{aligned}&\begin{aligned}&\sin \angle ACB=\displaystyle \frac{a}{c}\\ &\cos \angle ACB=\displaystyle \frac{b}{c}\\ &\tan \angle ACB=\displaystyle \frac{a}{b}=\displaystyle \frac{\sin \angle ACB}{\cos \angle ACB}\\ &\csc \angle ACB=\displaystyle \frac{c}{a}\\ &\sec \angle ACB=\displaystyle \frac{c}{b}\\ &\cot \angle ACB=\displaystyle \frac{b}{a}=\displaystyle \frac{\cos \angle ACB}{\sin \angle ACB} \end{aligned} \end{array}$

$\color{purple}\textrm{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\textrm{Perhatikan lagi gambar di poin c di atas}\\ &\begin{array}{|c|l|}\hline 1.&\textrm{Rumus saat dibagi dengan}\: \: c^{2}\\ &\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=\displaystyle \frac{c^{2}}{c^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=1\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{c} \right )^{2}+\left ( \displaystyle \frac{b}{c} \right )^{2}=1\\ &\color{blue}\sin ^{2}\angle ACB+\cos ^{2}\angle ACB=1\\\hline 2&\textrm{Rumus saat dibagi dengan}\: \: b^{2}\\ &\displaystyle \frac{a^{2}}{b^{2}}+\displaystyle \frac{b^{2}}{b^{2}}=\displaystyle \frac{c^{2}}{b^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{b^{2}}+1=\displaystyle \frac{c^{2}}{b^{2}}\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{b} \right )^{2}+1=\left ( \displaystyle \frac{c}{b} \right )^{2}\\ &\color{blue}\tan ^{2}\angle ACB+1=\sec ^{2}\angle ACB\\\hline 3&\textrm{Rumus saat dibagi dengan}\: \: a^{2}\\ &\displaystyle \frac{a^{2}}{a^{2}}+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\Leftrightarrow \color{red}1+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\\&\\ &\textrm{menjadi}\: \: \: 1+\left ( \displaystyle \frac{b}{a} \right )^{2}=\left ( \displaystyle \frac{c}{a} \right )^{2}\\ &\color{blue}1+\cot ^{2}\angle ACB=\csc ^{2}\angle ACB\\\hline \end{array} \end{aligned}$

$\color{purple}\textrm{c. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|c|c|c|c|c|c|c|}\hline \alpha &0^{\circ}&30&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textrm{TD}&0\\\hline \end{array}$.

$\begin{aligned}&\color{blue}\textrm{d. Macam-Macam Identitas Trigonometri Dasar}\\ &1.\quad \csc \alpha =\displaystyle \frac{1}{\sin \alpha }\qquad\qquad 5.\quad \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &2.\quad \sec \alpha =\displaystyle \frac{1}{\cos \alpha }\qquad\qquad 6.\quad \tan^{2} \alpha +1=\sec ^{2}\alpha \\ &3.\quad \cot \alpha =\displaystyle \frac{1}{\tan \alpha }\qquad\qquad 7.\quad \cot^{2} \alpha +1=\csc ^{2}\alpha \\ &4.\quad \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha }\qquad\qquad 8.\quad \sin^{2} \alpha +\cos ^{2}=1\\ \end{aligned}$.

$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad\tan \alpha =\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin ^{2}\alpha }\\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{\cos^{2} \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin^{2} \alpha }\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta =\cos \beta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta &=\displaystyle \frac{1}{\tan \beta }\times \sin \beta \\ &=\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta \\ &=\cos \beta \qquad\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } =1+\sin \gamma \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } &=\displaystyle \frac{1-\sin^{2} \gamma }{1-\sin \gamma }\\ &=\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }\\ &=1+\sin \gamma \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } =\cos ^{2}\theta -\sin ^{2}\theta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } &=\displaystyle \frac{1-\tan ^{2}\theta }{\sec ^{2}\theta }=\displaystyle \frac{1-\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }}\\ &=\displaystyle \frac{\displaystyle \frac{\cos ^{2}\theta -\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }} \\ &=\cos ^{2}\theta -\sin ^{2}\theta\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \cos ^{4}\alpha -\sin ^{4}\alpha =1-2\sin ^{2}\alpha \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\cos ^{4}\alpha -\sin ^{4}\alpha &=\left ( \cos ^{2}\alpha \right )^{2} -\left (\sin ^{2}\alpha \right )^{2}\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\left ( \cos ^{2}\alpha +\sin ^{2}\alpha \right )\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\times 1\\ &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \\ &=1-2\sin ^{2}\alpha \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta } =-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta }&=\displaystyle \frac{\sin \beta \left ( \displaystyle \frac{1}{\cos \beta } \right ) }{\sin ^{2}\beta -\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta } } \\ &=\displaystyle \frac{\left ( \displaystyle \frac{\sin \beta }{\cos \beta } \right )}{\sin ^{2}\beta \left ( 1-\displaystyle \frac{1}{\cos ^{2}\beta } \right )}\times \frac{\cos ^{2}\beta }{\cos ^{2}\beta }\\ &=\displaystyle \frac{\sin \beta \cos \beta }{\sin ^{2}\beta \left ( \cos ^{2}\beta -1 \right )}\\ &=\displaystyle \frac{\cos \beta }{\sin \beta \left ( -\sin ^{2}\beta \right )}\\ &=-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa} \\ &\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\displaystyle \frac{1}{\tan ^{2}x}}}=\sec ^{2}x \\\\&\color{blue}\textrm{Bukti}:\\&\begin{aligned}&\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\displaystyle \frac{1}{\tan ^{2}x}}}=\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\cot ^{2}x}}\\ &=\displaystyle \frac{1}{1-\displaystyle \frac{1}{\csc ^{2}x}}=\displaystyle \frac{1}{1-\sin ^{2}x}=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x\qquad \blacksquare \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Noormandiri, B. K. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Contoh 6 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 26.&\textrm{Diketahui lingkaran-lingkaran}\\ & x^{2}+y^{2}-2x+3y+k=0\: \: \textrm{dan}\: \\ &x^{2}+y^{2}+8x-6y-7=0\: \: \textrm{saling}\\ &\textrm{berpotongan ortogonal saat}\: \: k=\: ....\\ &\textrm{a}.\quad \color{red}-10\\ &\textrm{b}.\quad -3\\ &\textrm{c}.\quad 1\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 8\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}-2x+3y+k=0&\begin{cases} P_{1} &=\left ( 1,-\displaystyle \frac{3}{2} \right ) \\ r_{1} & = \sqrt{\displaystyle \frac{13-4k}{4}} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv x^{2}+y^{2}+8x-6y-7=0 \end{aligned}&\begin{cases} P_{2} &=\left ( -4,3 \right ) \\ r_{2} & = \sqrt{32} \end{cases}\\\hline \end{array} \\ &\textrm{Syarat dua lingkaran berpotongan ortogonal}\\ &\begin{aligned}&\left (P_{1}P_{2} \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( 1+4 \right )^{2}+\left ( -\displaystyle \frac{3}{2}-3 \right )^{2}=\left ( \sqrt{\displaystyle \frac{13-4k}{4}} \right )^{2}+\sqrt{32}^{2}\\ &\Leftrightarrow \: 25+\displaystyle \frac{81}{4}=\displaystyle \frac{13-4k}{4}+32\\ &\Leftrightarrow \: 100+81=13-4k+128\\ &\Leftrightarrow \: k=-10 \end{aligned} \\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Persamaan lingkaran yang berpotongan}\\ &\textrm{lingkaran lain}\: \: x^{2}+y^{2}+2x+y-11=0\\ &\textrm{secara tegak lurus dan melalui}\: \: (4,3)\: \: \textrm{serta}\\ &\textrm{pusatnya pada}\: \: 9x+4y=37\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-10x+4y+3=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x+10y+6=0\\ &\textrm{c}.\quad x^{2}+y^{2}+4x-8y+7=0\\ &\textrm{d}.\quad x^{2}+y^{2}+6x+y+5=0\\ &\textrm{e}.\quad x^{2}+y^{2}+12x+6y+5=0\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x+y-11=0&\begin{cases} P_{1} &=\left ( -1,-\displaystyle \frac{1}{2} \right ) \\ r_{1} & = \sqrt{\displaystyle \frac{49}{4}}=\displaystyle \frac{7}{2} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv (x-a)^{2}+(y-b)^{2}=r^{2} \end{aligned}&\begin{cases} P_{2} &=\left ( a,b \right ) \\ r_{2} & = r \end{cases}\\\hline \end{array}\\ &\textrm{Karena berpotongan tegak lurus, maka}\\ &\begin{aligned}&\left (P_{1}P_{2} \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( -1-a \right )^{2}+\left ( -\displaystyle \frac{1}{2}-b \right )^{2}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+2a+1+b^{2}+b+\displaystyle \frac{1}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow \color{blue}a^{2}+b^{2}+2a+b+\displaystyle \frac{5}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+b^{2}+2a+b-11=r^{2}\: .......(1)\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}&\textrm{Lingkaran}\: \: L_{2}\: \: \textrm{melalui titik}\: \: (4,3), \textrm{artinya}\\ &\textrm{bahwa}\: :\: (4-a)^{2}+(3-b)^{2}=r^{2}\\ &\Leftrightarrow a^{2}-8a+16+b^{2}-6b+9=r^{2}\\ &\Leftrightarrow a^{2}+b^{2}-8a-6b+25=r^{2}\: .......(2)\\ &\textrm{Pusat lingkaran}\: \: L_{2}\: \: \textrm{melalui garis}\: \: 9x+4y=37\\ &\textrm{artinya}:\: 9a+4b=37\: ...............(3)\\ \end{aligned}\\ &\begin{aligned}&\textrm{Dengan eliminasi}\: 1\: \&\: 2\: \: \textrm{dapat diperoleh}:\\ &\begin{array}{rll} a^{2}+b^{2}-8a-6b+25&=r^{2}&\\ a^{2}+b^{2}+2a+b-11&=r^{2}&-\\\hline -10a-7b+36&=0&\textrm{atau}\\ 10a+7b&=36&......(4) \end{array}\\ &\textrm{Dari persamaan}\: 3\: \&\: 4\: \: \textrm{dapat diperoleh}:\\ & \end{aligned}\\ &\begin{array}{rll} 10a+7b&=36&(\times 4)\\ 9a+4b&=37&(\times 7)\\\hline 40a+28b&=144&\\ 63a+28b&=259&\\\hline -23a\: \quad\quad&=-115&\\ a&=\displaystyle \frac{-115}{-23}&=5\\ 10(5)+7b&=36&\\ 7b&=-14\\ b&=-2 \end{array}\\ &\textrm{Adapun langkah berikutnya}\\ &\begin{aligned}&L_{2}\equiv (4-a)^{2}+(3-b)^{2}=r^{2}\\ &L_{2}\equiv (4-5)^{2}+(3+2)^{2}=r^{2}\\ &L_{2}\equiv r^{2}=25+1=26\\ &\textrm{Sehingga},\: L_{2}\equiv (x-5)^{2}+(y+2)^{2}=26\\ &\Leftrightarrow x^{2}+y^{2}-10x+4y+25+4-26=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-10x+4y+3=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.

Jika diperjelas dengan tambahan garis 9x+4y=37

$\begin{array}{ll}\\ 28.&\textrm{Diketahui lingkaran pertama berpusat di}\: \: (1,2)\\ &\textrm{dan menyinggung garis}\: \: 3x-4y+10=0.\\ &\textrm{Jika ada lingkaran kedua dengan pusat}\: \: (4,6)\\ &\textrm{dan menyinggung lingkaran yang pertama},\\ &\textrm{maka persamaan lingkaran yang kedua}\\ &\textrm{tersebut adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-8x-12y+48=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x-12y+43=0\\ &\textrm{c}.\quad \color{red}x^{2}+y^{2}-8x-12y+36=0\\ &\textrm{d}.\quad x^{2}+y^{2}-8x-12y+27=0\\ &\textrm{e}.\quad x^{2}+y^{2}-8x-12y+16=0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa kedua lingkaran saling}\\ &\color{blue}\textrm{bersinggungan di luar},\: \color{black}\textrm{maka}\\ &\begin{aligned}r_{1}+r_{2}&=P_{1}P_{2}\\ &=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}\\ &=\sqrt{(1-4)^{2}+(2-6)^{2}}\\ &=\sqrt{3^{2}+4^{2}}=\sqrt{5^{2}}=5 \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}r_{\textrm{pertama}}&=\left |\displaystyle \frac{3(1)-4(2)+10}{\sqrt{3^{2}+4^{2}}} \right |\\ &=\left | \displaystyle \frac{3-8+10}{\sqrt{5^{2}}} \right |=\left | \displaystyle \frac{5}{5} \right |=\left | 1 \right |=1\\ \textrm{sehingga} &\\ r_{\textrm{kedua}}&=5-r_{\textrm{pertama}}=5-1=4 \end{aligned}\\ &\textrm{maka persamaan lingkaran keduanya adalah}:\\ &\begin{aligned}&(x-4)^{2}+(y-6)^{2}=4^{2}\\ &\Leftrightarrow x^{2}-8x+16+y^{2}-12y+36=16\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-8x-12y+36=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Garis kuasa (tali busur sekutu)}\\ &\textrm{dari lingkaran}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 3x+4y+9=0\\ &\textrm{b}.\quad 3x-4y-8=0\\ &\textrm{c}.\quad 3x-4y+7=0\\ &\textrm{d}.\quad 3x+4y-7=0\\ &\textrm{e}.\quad \color{red}3x+4y-6=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0,\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}+6x-4y-12\\ &-(x^{2}+y^{2}-12y)=0\\ &\Leftrightarrow 6x+8y-12=0\\ &\Leftrightarrow \color{red}3x+4y-6=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Jika dua lingkaran}\\ & x^{2}+y^{2}=9\: \: \textrm{dan}\\ &x^{2}+y^{2}-4y+2y+3=0\: \: \textrm{yang}\\ &\textrm{berpotongan di}\: \: (x_{1},y_{1})\: \: \textrm{dan}\: \: (x_{2},y_{2}),\\ &\textrm{maka nilai}\: \: 5(x_{1}+x_{2})\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}24\\ &\textrm{b}.\quad 26\\ &\textrm{c}.\quad 28\\ &\textrm{d}.\quad 30\\ &\textrm{e}.\quad 32\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}-9=0\: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+y^{2}-4x+2y+3\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}-9\\ &-(x^{2}+y^{2}-4y+2y+3)=0\\ &\Leftrightarrow 4x-2y-12=0\\ &\Leftrightarrow 2x-y-6=0\\ &\Leftrightarrow y=6-2x \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}&x^{2}+y^{2}-9=0\\ &\Leftrightarrow x^{2}+(6-2x)^{2}-9=0\\ &\Leftrightarrow x^{2}+36-24x+4x^{2}-9=0\\ &\Leftrightarrow 5x^{2}-24x+27=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{576-540}}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow \quad x_{1}=3\: \: \textrm{atau}\: \: x_{2}=1,8\\ &\textrm{maka}\: \: 5(x_{1}+x_{2})=5\left ( 3+1,8 \right )=\color{red}24 \end{aligned} \end{array}$.